Find maximum and minimum values of f(x,y) = xy on the circle g(x,y)= x^2 +y^2 −1=0 .


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Question Number 122695 by liberty last updated on 19/Nov/20

$F i n d m a x i m u m a n d m i n i m u m v a l u e s$ $o f f (x, y) = x y o n t h e c i r c l e g (x, y) =$ $x^{2} + y^{2} - 1 = 0 .$
	Answered by john santu last updated on 19/Nov/20
	$By Lagrange multiplier ▽f(p) = λ ▽g(p) ,where P is any point (((∂f/∂x)),((∂f/∂y)) ) = λ (((∂g/∂x)),((∂g/∂y)) ) ((y),(x) ) = (((2λx)),((2λy)) ) → { ((λ=(y/(2x)))),((λ=(x/(2y)))) :} ⇒ (y/(2x)) = (x/(2y)) ⇒x^2 =y^2 ∧x^2 +y^2 =1 this gives four solutions { ((((1/( (√2))),(1/( (√2)))) , ((1/( (√2))),−(1/( (√2)))))),(((−(1/( (√2))),(1/( (√2)))) ,(−(1/( (√2))),−(1/( (√2)))))) :} Evaluating f at these four points we get max value is (1/2) and min value is −(1/2).$
	$B y L a g r a n g e m u l t i p l i e r$ $▽ f (p) = λ ▽ g (p), w h e r e P i s a n y p o i n t$ $(\begin{matrix} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \end{matrix}) = λ (\begin{matrix} \frac{\partial g}{\partial x} \\ \frac{\partial g}{\partial y} \end{matrix})$ $(\begin{matrix} y \\ x \end{matrix}) = (\begin{matrix} 2 λ x \\ 2 λ y \end{matrix}) \to {\begin{cases} λ = \frac{y}{2 x} \\ λ = \frac{x}{2 y} \end{cases}$ $\Rightarrow \frac{y}{2 x} = \frac{x}{2 y} \Rightarrow x^{2} = y^{2} \land x^{2} + y^{2} = 1$ $t h i s g i v e s f o u r s o l u t i o n s$ ${\begin{cases} (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}), (\frac{1}{\sqrt{2}}, - \frac{1}{\sqrt{2}}) \\ (- \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}), (- \frac{1}{\sqrt{2}}, - \frac{1}{\sqrt{2}}) \end{cases}$ $E v a l u a t i n g f a t t h e s e f o u r p o i n t s$ $w e g e t m a x v a l u e i s \frac{1}{2} a n d m i n v a l u e$ $i s - \frac{1}{2} .$
	Answered by mr W last updated on 19/Nov/20

	$P = x y$ $y = \frac{P}{x}$ $x^{2} + \frac{P^{2}}{x^{2}} - 1 = 0$ $2 x + \frac{2 {P P}^{'}}{x^{2}} - \frac{2 P^{2}}{x^{3}} = 0$ $P^{'} = 0$ $x - \frac{P^{2}}{x^{3}} = 0$ $x - \frac{x^{2} y^{2}}{x^{3}} = 0$ $x^{2} = y^{2} = \frac{1}{2}$ $\Rightarrow x = \pm \frac{1}{\sqrt{2}}, y = \pm \frac{1}{\sqrt{2}}$ ${(x y)}_{m a x} = \frac{1}{2}$ ${(x y)}_{m i n} = - \frac{1}{2}$
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